Question

我目前在我的网站上设置了 AJAX 脚本，该脚本根据下拉菜单[id="afl_player_ID"]中的选择来更新表格。

如果[id="afl_player_ID"]或[id="afl_stat_ID"]被更改，我想做的就是运行相同的脚本。

如何将此功能添加到此代码中？

<script type="text/javascript">
jQuery(document).ready(function ($) {
    var valueCheck;
    jQuery("#afl_player_ID").on("change", function () {
        player_ID = $("#afl_player_ID").val();
        stat_ID = $("#afl_stat_ID").val();
        jQuery.ajax({
            type: "POST",
            url: "/wp-admin/admin-ajax.php",
            data: {
                action: "call_advanced_player_basic_gamelog",
                player_ID: player_ID,
                stat_ID: stat_ID
            },
            success: function (output) {
                jQuery("#advanced_player_basic_gamelog").html(output);
            }
        });
    }).change();
});
</script>

Answer 1

使用@Farooq Khan在评论中建议的以下代码

<script type="text/javascript">
jQuery(document).ready( function($) {
    var valueCheck;
    jQuery('#afl_player_ID,#afl_stat_ID').on( 'change', function () {
         player_ID = $('#afl_player_ID').val();
         stat_ID = $('#afl_stat_ID').val();
     jQuery.ajax({
        type: "POST",
        url: "/wp-admin/admin-ajax.php",
        data: {
            action: 'call_advanced_player_basic_gamelog',
            player_ID: player_ID,
            stat_ID: stat_ID,
        },
         success:function(output){
             jQuery('#advanced_player_basic_gamelog').html( output );
         }
     });
    }).change();
});
</script>

Javascript-添加更多on.change选项

1 个答案: